Question

Find the length of the hypotenuse of the right triangle MPQ, if tanM = square root of 6/ 5 and angle P is the right angle

Answer 1

Given:

A right angle triangle ΔMPQ where angle P is the right angle.

$\mathbf{if\ \tan M=\sqrt{\dfrac{6}{5}}}$

To Find:

What is value of hypotenuse MQ?

Solution:

$\mathbf{\tan M=\sqrt{\dfrac{6}{5}}=\dfrac{\sqrt{6}\ X}{\sqrt{5}\ X}}$

Means:

$\mathbf{PQ=\sqrt{6}\ X}$       ...1)

$\mathbf{MP=\sqrt{5}\ X}$      ...2)

From Pythagoras' rules:

MQ² = MP² + PQ²

On putting respective value in above equation:

$\mathbf{MQ^{2}=(\sqrt{5}\ X)^{2}+(\sqrt{6}\ X)^{2}}$

$\mathbf{MQ^{2}=5\ X^{2}+6\ X^{2}}$

$\mathbf{MQ^{2}=11\ X^{2}}$

So:

Hypotenuse = MQ = √11 X

Means relation between hypotenuse, perpendicular and base is √11:√6:√5

Answer 2

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Means relation between hypotenuse, perpendicular and base is √11:√6:√5